Question: $3qr - 4qs - 4q - 9 = -r + 2$ Solve for $q$.
Combine constant terms on the right. $3qr - 4qs - 4q - {9} = -r + {2}$ $3qr - 4qs - 4q = -r + {11}$ Notice that all the terms on the left-hand side of the equation have $q$ in them. $3{q}r - 4{q}s - 4{q} = -r + 11$ Factor out the $q$ ${q} \cdot \left( 3r - 4s - 4 \right) = -r + 11$ Isolate the $q$ $q \cdot \left( {3r - 4s - 4} \right) = -r + 11$ $q = \dfrac{ -r + 11 }{ {3r - 4s - 4} }$ We can simplify this by multiplying the top and bottom by $-1$. $q= \dfrac{r - 11}{-3r + 4s + 4}$